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Math 718: Lie Algebras

Taught by Eric Sommers; Notes by Patrick Lei

University of Massachusetts, Amherst

Fall 2019

Abstract

The goal of the course is the classification of semisimple Lie algebras over the complexnumbers and an introduction to their representation theory. The classification is essentially thesame as that for simply-connected compact Lie groups, which are central objects in mathematicsand physics. Topics covered include representations of sl(2), Jordan decomposition, structure the-orems, Weyl groups, roots systems, Dynkin diagrams, complete reducibility, finite-dimensionalrepresentations and their characters formulas.

Contents

1 Lecture 1 (Sep 3) 31.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Basic Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Lecture 2 (Sep 5) 42.1 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Basic Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 More Basic Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Lecture 3 (Sep 10) 63.1 Nilpotent Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 Lecture 4 (Sep 12) 74.1 More Nilpotent Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74.2 Solvable and Semisimple Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

5 Lecture 5 (Sep 17) 85.1 Jordan Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

6 Lecture 6 (Sep 19) 96.1 Proof of Theorem 48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

7 Lecture 7 (Sep 24) 107.1 Some Classical Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

8 Lecture 8 (Sep 26) 118.1 Representations of Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118.2 Jordan decomposition for Semisimple Lie Algebras . . . . . . . . . . . . . . . . . . . 11

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Patrick Lei University of Massachusetts, Amherst Fall 2019

9 Lecture 9 (Oct 1) 129.1 Jordan Decomposition Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129.2 Representations of sl2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

10 Lecture 10 (Oct 3) 1410.1 Representation Theory of sl2 Continued . . . . . . . . . . . . . . . . . . . . . . . . . . 1410.2 Maximal Tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

11 Lecture 11 (Oct 08) 1511.1 Tori in sln . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1511.2 The General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

12 Lecture 12 (Oct 10) 17

13 Lecture 13 (Oct 17) 1813.1 An Example: The symplectic Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . 19

14 Lecture 14 (Oct 22) 1914.1 Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1914.2 Classifying Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

15 Lecture 15 (Oct 24) 2015.1 Weyl Groups continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20



18 Lecture 18 (Nov 5) 23






24 Lecture 24 (Dec 3) 29



Disclaimer: Any errors in these notes are mine and not the instructor’s. In addition, these notesare picture-free (but will contain commutative diagrams) and may not be exactly the same as

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Math 718 Lie Algebras Lecture Notes

the lectures (I tend to leave out boring computations, streamline certain things, and use categorytheory).

1 Lecture 1 (Sep 3)

We will be using the book by Humphreys in this course, but we will not be following the book tooclosely. Homework will be approximately weekly, with the first due in two weeks.

1.1 Overview First we will discuss symmetry. Recall the platonic structures and their dualstructure. We can consider the finite reflection group that preserves the object, which leads us tothe following question:

Question 1. What are the finite reflection groups?

In the case of Lie Algebras, we will need our groups to preserve integer lattices (root systems), sowe specialize the question:

Question 2. What are the finite reflection groups that can be represented by integer matrices?

In two dimensions, only the dihedral group for the triangle, square, and hexagon are representableby integer matrices. We will reduce the classification of semisimple Lie algebras to the abovequestion.

Intimately related to Lie algebras are Lie groups, which are just group objects in Diff. We willnot need Lie groups in this course. However, it is worth mentioning that Lie algebras arise as thetangent space at the identity of a Lie group G. We can also pass to algebraic groups, which arejust group objects in a category of varieties. The advantage of this approach is that we can workover arbitrary fields.

1.2 Basic Notions We will now define Lie algebras. First recall that a k-algebra A is a k-vectorspace A with a bilinear map A⊗A→ A. If A is unital, we call it central. If A is associative, thenA is a monoid in Vectk.

Example 3. An example of an associative algebra is a matrix algebra Mn(k).

Definition 4. A Lie Algebra g is a k-algebra with a multiplication

[−,−] : g∧ g→ g

satisfying the Jacobi identity:

[x, [y, z]] + [y, [z, x]] + [z, [x,y]] = 0.

Example 5. From any associative algebra A, we can build a Lie algebra using the commutator asthe Lie bracket. In particular, Mn(k) can be turned into a Lie algebra gln.

The notion of a morphism of Lie algebras is entirely analogous to the case of rings, so we candefine Lie subalgebras.

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Example 6. Some subalgebras of gln are the set of diagonal matrices, which is abelian, the set ofstrictly upper triangular matrices, and the set of upper triangular matrices b.1.

Definition 7. A Lie algebra is abelian if [−,−] = 0.

Example 8. An important subalgbebra of gln is the set sln of matrices of trace 0.

Remark 9. sln is what is called a simple Lie algebra whose Weyl group is Sn.

2 Lecture 2 (Sep 5)

2.1 Derivations First we discuss derivations.

Definition 10. Let A be a k-algebra. Then δ ∈ EndA is a derivation if

δ(ab) = δ(a)b+ aδ(b)

for all a,b ∈ A.

We will denote the derivations of g by Der g.

Remark 11. Der g is a Lie subalgebra of End g.

The Jacobi identity is related to derivations. Recall the adjoint representation of a Lie algebra g.

Definition 12. For x ∈ g, write adx for the adjoint map

adx : y 7→ [x,y].

Now observe that

[x, [y, z]] = −[y, [x, z]] − [z, [x,y]]= −[z, [x,y]] − [y, [z, x]]= [[x,y], z] + [y, [x, z]],

so we see that adx is a derivation.

2.2 Basic Examples We will now expand our set of examples of Lie algebras.

Example 13. The orthogonal group O(n) is the set of matrices that preserve a Hermitian innerproduct. Equivalently, O(n) is defined by the equation XtX = I.

Remark 14. On the Lie group side, we may define subgroups of GLn via XtJX = J for invertiblematrices J. For Lie algebras, this differentiates to XtJ+ JX = 0.

Example 15. For k = C, the orthogonal Lie algebra is denoted by o(n) and is given by J = I. Wemay also define the symplectic Lie algebra sp(2n) by

J =

(0 I−I 0

).

1For Borel

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Remark 16. The orthogonal groups behave differently in even and ood dimension. In fact, they aregiven by different classes of Dynkin diagrams.

Remark 17. The four infinite families of simple Lie algebras are sp2n, o2n, o2n+1, sln. There are fiveother exceptional simple Lie algebras.

2.3 More Basic Notions We will now consider the basic structure of a Lie algebra.

Definition 18. The derived subalgebra [g, g] of g is the span of all elements of the form [x,y].

This is clearly a subalgebra of g, but in fact it is an ideal.

Definition 19. An ideal I of g is a subalgebra satisfying [g, I] ⊂ I.

Example 20. The center of a Lie algebra is an ideal. To see this, note that [A,X] = 0 for all A ∈ gand X ∈ Z(g). Also, it is easy to see that the derived subalgebra is an ideal.

Definition 21. A Lie algebra g is nilpotent if the lower central series

g, [g, g], [g, [g, g]], . . .

eventually becomes 0.

Example 22. Clearly all abelian Lie algebras are nilpotent because [g, g] = 0.

Definition 23. A Lie algebra g is solvable if the upper central series

g, [g, g], [[g, g], [g, g]], . . .

eventually becomes 0.

Example 24. We note that all nilpotent Lie algebras are solvable because the upper central seriesis contained in the lower central series.

Example 25. The set of all upper triangular matrices is solvable, and the set of strictly uppertriangular matrices is nilpotent.

Definition 26. The radical Rad g is the maximal solvable ideal in g.

Definition 27. A Lie algebra g is semisimple if Rad g = 0.

Definition 28. A Lie algebra g is simple if it has no nontrivial ideals and dimk g 6= 1.

Example 29. The simplest example of a simple Lie algebra is sl2.

Homomorphisms and quotients are defined entirely analogously to the case of rings or groups.Checking this is well-defined is left to the reader.

Remark 30. The first isomorphism theorem holds for Lie algebras.

Recall that a representation of a finite group G is a morphism G→ GL(V) for some vector spaceV . Alternately, this is a functor BG→ Vectk.

Definition 31. A representation of a Lie algebra g is a morphism g→ EndV for some vector spaceV . Alternatively, we can define this as a g-module.2

2Better, we can define this as a representation of the universal enveloping algebra, which allows us to define representa-tions as functors Ug→ Vectk.

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3 Lecture 3 (Sep 10)

Recall the adjoint representation of a Lie algebra g. This is a very important representation. For asimple Lie algebra, ad is an isomorphism onto its image. In general, we will write ad g for theimage of the adjoint map. In general, the kernel of ad is the center.

Remark 32. When we define the E8 algebra, which has dimension 248, we will have a 248-dimensional subalgebra of a 2482-dimensional Lie algebra.

Remark 33. From now on, we will only consider finite-dimensional Lie algebras.

3.1 Nilpotent Lie Algebras

Notation 34. We will denote the set of all strictly upper triangular matrices by nn(k).

Eric then proceeded to spend time computing products and Lie brackets of matrices of the formeij.3 It is then clear that n is nilpotent because bracketing shifts nonzero entries towards the topright corner.

We now discuss some properties of nilpotent Lie algebras.

Proposition 35. The following hold for nonzero nilpotent Lie algebras:

1. Subalgebras and quotients of nilpotent Lie algebras are nilpotent.

2. If g/Z(g) is nilpotent, then so is g.

3. If g is nilpotent, it has nonzero center.

Proof of these is easy and is omitted.

Remark 36. The adjoint representation of a nilpotent Lie algebra consists entirely of nilpotentmatrices.

Theorem 37 (Engel). If adx is nilpotent for all x ∈ g, then g is a nilpotent Lie algebra.

First, we need the following lemma:

Lemma 38. Suppose g ⊂ gl(V) consists of nilpotent endomorphisms. Then there exists v ∈ V \ 0 suchthat zv = 0 for all z ∈ g.

Proof of Engel’s Theorem. Take the adjoint representation of g. By hypothesis, this consists ofnilpotent endomorphisms. By Lemma 38, then we have a common eigenvector, which is in thecenter. Thus ad g has smaller dimension than g.

To complete the proof with induction, we need to show that nilpotent matrices are sent to milpotentmatrices by ad. However, this is clear from the definition of commutator and nilpotence of theoriginal matrix. Now, by induction, ad g is a nilpotent Lie algebra. Because ad g ' g/Z(g), we usepart 2 of Proposition 35 to conclude that g is nilpotent.

3This demonstrates the axiom that mathematicians are terrible at computing things.

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Proof of Lemma 38. If K ⊂ g is a proper maximal subalgebra, then the adjoint representation of Kdescends to g/K. Then ad z is nilpotent for z ∈ K. Therefore the image of K in gl(g/K) consists ofnilpotent endomorphisms. By induction on dimension, there is a common engenvector w for K.

Now consider w = l+ K. Then K+ k · l must generate all of g with K an ideal of g. ViewingK ⊂ g ⊂ gl(V), induction gives us an eigenvector in V for K. Define W to be the subspace spannedby the eigenvectors of K. Then it is easy to check that l preserves W. Now choose v ∈ W aneigenvector of l, and then K+ k · l annihlates v. Thus L annihlates v.


We began class ( 30 minutes) by finishing the class of Lemma 38 from last time.

4.1 More Nilpotent Lie Algebras

Corollary 39. Let L ⊂ gl(V) consist of nilpotent endomorphisms. Then there exists a basis of V such thatL is strictly upper triangular in this basis.4

Proof. By Lemma 38, there exists a nonzero v1 ∈ V such that L annihlates v1. Then L is nilpotenton V/〈v1〉. Now induct.

Corollary 40. If L is a nilpotent Lie algebra, and 0 6= K ⊂ L is an ideal, then K∩Z(L) 6= 0.

4.2 Solvable and Semisimple Lie Algebras We now discuss solvable Lie algebras. We knowalready that nilpotent implies solvable. Recall that RadL is defined to be the maximal solvableideal of L. This is unique and contains all solvable ideals.

Definition 41. L is semisimple if RadL = 0.5

Proposition 42. Simple implies semisimple.

Proof. [L,L] = L because L is simple, so L cannot be solvable. Therefore there are no nonzerosolvable ideals.

Now we will define a symmetric bilinear form κ : S2(L)→ k given by

κ(X, Y) = tr(adX adY)

known as the Killing form.6

From now on, we will work in an algebraically closed field of characteristic 0.

Theorem 43 (Killing). L is semisimple if and only if κ is non-degenerate.

Theorem 44 (Lie). If L ⊂ gl(V) is solvable, then there exists a common eigenvector for L.

4This is the same as L fixing a flag F. In fact, L.Fi ⊂ Fi−1.5These are the main object of study in this course. Also, these will turn out to be sums of simple objects.6This actually turns out to be a multiple of the normal trace form for the classical Lie algebras.

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We concluded class with a linear algebra review. Specifically, observe the following:

1. Diagonalizable is equivalent to semisimple;

2. If [x,y] = 0, then:

• x,y nilpotent implies x+ y nilpotent;

• x,y semisimple implies x+ y semisimple.

3. If W ⊂ V and x(W) ⊂W, then:

• x|W is semisimple if x is semisimple.

• x|W is nilpotent if x is nilpotent.


5.1 Jordan Decomposition Class began with a discussion of the Jordan canonical form. Inparticular, recall:

Proposition 45. x ∈ EndV can be written x = xs + xn where xs is semisimple and xn is nilpotent, and[xs, xn] = 0. In particular, they are both polynomials in x with constant term 0. Finally, x(B) ⊂ A if andonly if xs(B), xn(B) ⊂ A for subspaces A ⊂ B ⊂ V .

Proof of this can be found in any standard algebra text. In addition, the decomposition is unique.

Now we would like to have the notion of Jordan decomposition in an arbitrary Lie algebra. Thisonly has nice properties for semisimpe Lie algebras. In particular, we want:

1. If L ⊂ gl(V), then the Jordan decomposition agrees with that of gl(V).

2. The Jordan decomposition respects morphisms.

Definition 46. For a semisimple Lie algebra, an abstract Jordan decomposition is a Jordan decompo-sition of ad x = ad xs + ad xn

Corollary 47. Let x ∈ gl(V) with x = xs + xn. Then (ad x)s = ad xs and (ad x)n = ad xn.

Proof. Note that if x is nilpotent, then ad x is also nilpotent. In addition, if x is semisimple, choosea basis of V such that x is diagonal. Then it is easy to see that ad x is semisimple. Finally, notethat because xs, xn commute, so do their adjoint operators.

Recall the symmetric bilinear form 〈−,−〉 : Mn ⊗Mn → F given by A⊗ B 7→ tr(AB). We canverify that this is associative with respect to the Lie bracket:

〈[x.y], z〉 = 〈x, [y, z]〉.

Also, it is easy to verify that this bilinear form is nondegenerate. Now recall the Killing form on Lgiven by

L⊗ L ad⊗ ad−−−−→ gl(L)⊗ gl(L)〈−,−〉−−−−→ F.

This is an associative symmetric bilinear form, but in general, it is degenerate.

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Next time, we will show that

Theorem 48. L is semisimple if and only if the Killing form is nondegenerate.


6.1 Proof of Theorem 48 Today we begin with the proof of Theorem 48.

Lemma 49. Let A ⊂ B be subspaces of gl(V). Define

M := x ∈ gl(V) | [x,B] ⊂ A.

Suppose x ∈M satisfies tr(xy) = 0 for all y ∈M. Then x is nilpotent.

Proof of this fact is in Humphreys and uses the Jordan canonical form and the fact that our fieldhas characteristic 0.

Theorem 50 (Cartan’s Criterion). Let L ⊂ gl(V). Suppose tr(xy) = 0 for all x ∈ [L,L],y ∈ L. Then L issolvable.

Proof. Note that it is enough to show that x ∈ [L,L] is nilpotent. We will apply Lemma 49 toA = [L,L],B = L. Note that L ⊂M. We know that [L,L] is generated by [y, z] for all y, z ∈ L. Letm ∈M. We know that

tr([y, z]m) = tr(y[z,m]) = tr([z,m]y) = 0

by hypothesis. Because [y, z] is a generator of [L,L], we know that tr(xm) = 0 for all x ∈ [L,L],m ∈M. By Lemma 49, x is nilpotent.

Corollary 51. Let L be a Lie algebra such that tr(adx ady) = 0 for all x ∈ [L,L],y ∈ L. Then L is solvable.

Proof. adL is solvable by the theorem. Then L is solvable because L ' L/Z(L).

Lemma 52. Let I be an ideal of L and κI be the Killing form for I as a Lie algebra. Then κI = κ|I⊗I.

Proof of this amounts to the fact that the product respects the block form. To review some linearalgebra, given a symmetric bilinear form β : V ⊗ V → F, the radical is the kernel of the inducedmap x 7→ β(x,−). Then β is nondegenerate if the radical is trivial. Then note that if β is asymmetric bilinear associative form on a Lie algbera, the radical is an ideal.

Proof of Theorem 48. Let S be the radical of κ. Then it is easy to see that S is solvable by Corollary51. Thus S ⊂ RadL. Thus if L is semisimple, S = 0.

In the other direction, note that being semisimple is equivalent to having no nonzero abelian ideals.

Let I be a nonzero abelian ideal of L. Choose x ∈ I,y ∈ L. Then Lady−−−→ L

adx−−→ Iady−−−→ I

adx−−→ 0.Thus ady adx is nilpotent, so it has trace 0. Thus x ∈ S and thus I ⊂ S. In particular, if κ isnondegenerate, then S = 0, so every abelian ideal is zero. Therefore L is semisimple.

Theorem 53. L is semisimple if and only if there exists L1, . . . ,Lt ideals of L which are simple such thatL = L1 ⊕ · · · ⊕ Lt. Here the direct sum is just the coproduct.

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Also, note that every simple ideal in L coincides with some Li and that the direct sum decomposi-tion is unique.

Proof. Suppose L is semisimple. If L is simple, we are done. Otherwise, consider I ⊂ L a propernonzero ideal. Then it is easy to see that I⊥ is an ideal and that L = I⊕ I⊥. By induction, L is adirect sum of simple Lie algebras.

In the other direction, Note that [L,L] =⊕

[Li,Li] =⊕Li = L. Also note that RadL =

⊕RadLi =

0. This gives semisimplicity.


We began class by finishing the proof of Theorem 53. Everything is in the notes from last time.

Remark 54. If I is an ideal in L1 ⊕ L2, then I = I1 ⊕ I2.

Remark 55. Let L be semisimple and I ⊂ L an ideal. Then I = Li1 ⊕ · · · ⊕ Lik . In particular, the Liare the only simple ideals in L.

Thus our goal for this class is to classify the simple Lie algebras.

7.1 Some Classical Computations Now recall the definition of the orthogonal Lie algebra. Wewill take J to be the identity matrix. For n = 2, we have the set of matrices such that At +A = 0,which is one-dimensional. However, we will consider n = 4 with matrix

J =

0 0 0 10 0 1 00 1 0 01 0 0 0

.

Proposition 56. so4 ' sl2 ⊕ sl2.

Now we will compute the Killing form of sl2 explicitly. We will use the standard basis e,h, f. Bythe standard relations, we have:

ad e =

0 −2 00 0 10 0 0

; adh =

2 0 00 0 00 0 −2

; ad f =

0 0 0−1 0 00 2 0

.

Now we can see that

κ =

0 0 40 8 04 0 0

.

Now recall that a bilinear form is nondegenerate if and only if its matrix is nonsingular. Inaddition, a nondegenerate bilinear form gives an isomorphism to V∗. Thus we can construct adual basis in V .

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To motivate the dual basis, we can define a Casimir operator as follows: Choose a basis xifor L, and then choose a dual basis yi. Then in the universal enveloping algebra UL, considerΩ =

∑xiyi. Then Ω ∈ Z(UL).

Now recall Schur’s Lemma:

Lemma 57 (Schur). If V is an irreducible representation, then any nonzero equivariant endomorphism ofV is a scalar.


8.1 Representations of Lie Algebras Recall the definition of a representation of a Lie algebra L.Note that this is equivalent to the notion of a module over L. Then recall that a representation iscompletely reducible (or semisimple) if it is a direct sum of irreducible representations.

Theorem 58 (Weyl). If L is semisimple, then any representation of V is semisimple.

For finite groups, we have the analogous statement:

Theorem 59 (Maschke). Let G be a finite group. If chark - |G|, then any k-representation of G issemisimple.

Remark 60. The original proof of Weyl’s theorem was analytic and uses the unitary trick.

Note that if L is not semisimple, Weyl’s theorem can fail. For example, consider the standardtwo-dimensional representation of the set of upper triangular matrices. Then the x-axis is asubrepresentation, but there is no other common eigenvector.

Now we construct a representation structure on Hom(V ,W) by (x.f)(v) = x.f(v) − f(x.v), so L−mod is enriched over itself. In particular, the dual space of a representation is also a representation.Finally, we may construct a tensor product representation by x.(v⊗w) = (w.v)⊗w+ v⊗ (x.w).

Remark 61. The isomorphism V∗ ⊗W ' Hom(V ,W) holds in L-mod.

Remark 62. The tensor-hom adjunction holds for representations of Lie algebras, so L-mod is closedmonoidal.

Proof of Schur’s Lemma. Let 0 6= π : V →W be a morphism of irreducible representations. Considerthe kernel. For a nonzero morphism, this must be the source. In addition, the image must benonzero, so it is the entire target.

To show any nonzero endomorphism is a scalar, use algebraic closure to find an eigenvalue. Thenthe kernel of π− λI is nonzero, so it must be all of V .

Remark 63. When we prove Weyl’s theorem, we use β(x,y) = tr(φ(x)φ(y)). This is nondegeneratewhen the kernel is nonzero.

8.2 Jordan decomposition for Semisimple Lie Algebras For an arbitrary Lie algebra L, callx ∈ L semisimple if ad x is semisimple, and nilpotent if ad x is nilpotent. Then we can decomposead x = (ad x)s + (ad x)n. What we will show is that

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Lemma 64. If δ ∈ DerL, then δs, δn ∈ DerL.

Lemma 65. If L is semisimple, then DerL = adL.

Now we can form an abstract Jordan decomposition of any semisimple Lie algebra.

Theorem 66. Let φ be a representation of L and x = s+ n be the Jordan decomposition in L. Thenφ(x) = φ(x) +φ(n) is the Jordan decomposition in gl(V).

9 Lecture 9 (Oct 1)

9.1 Jordan Decomposition Continued

Proof of Lemma 65. Recall that if L is semisimple, then L ' adL. Then we know that κ on adL isnon-degenerate. Also, adL ⊂ DerL is an ideal. Now take (adL)⊥ and note that adL∩ (adL)⊥ = 0.Thus DerL = adL⊕ (adL)⊥. Then, by perpendicularity, the direct sub holds as Lie algebras.

We now recall that [δ, ad x] = ad δ(x). Now for δ ∈ (adL)⊥, we have

0 = [δ, ad x] = ad δ(x).

However, L is semisimple, so δ(x) = 0 for all x. The desired result follows..

As a consequence, we can describe an abstract Jordan decomposition on a semisimple Lie algebra.

Corollary 67 (Jordan Decomposition). Given x ∈ L, then the Jordan decomposition ad x = (ad x)s +(ad x)n restricts to

ad x = ad xs + ad xn

for some xs, xn ∈ L. In particular, x = xs + xn.

Now note that [xs, xn] = 0 because

ad[xs, xn] = [ad xs, ad xn] = 0

by the usual Jordan decomposition and because L is semisimple.

Remark 68 (Eric’s Research). Consider the group G = SL2(C) with Lie algebra g = sl2(C). Now Gacts on g by conjugation. We consider the following question:

What are the orbits of G on the set of nilpotent elements in g?

We find that there are two orbits of sl2 and that the set of nilpotents is a closed subvariety of g.

Theorem 69. Let L ⊂ gl(V) be semisimple. Then the usual Jordan decomposition in gl(V) agrees with theabstract Jordan decomposition in L.

Corollary 70. Let φ : L → gl(V) be any representation of L. Let x = xs + xn be the abstract Jordandecomposition. Then φ(x) = φ(xs) +φ(xn) is the usual Jordan decomposition in gl(V).

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9.2 Representations of sl2 We will classify irreducible representations of sl2. 7 Let V be somefinite-dimensional representation of sl2. Recall that sl2 has standard basis e,h, f with the usualrelations.

Now we note that h is semisimple in sl2, so φ(h) is semisimple for any representation φ. Then wecan take the eigenspace decomposition

V =⊕λ∈F

Vλ.

Lemma 71. If v ∈ Vλ, then e.v ∈ Vλ+2 and f.v ∈ Vλ−2.

Proof.h.(e.v) = [h.e].v+ e.(h.v) = (2e).v+ e.(λv) = (λ+ 2)(e.v)

andh.(f.v) = [h.f].v+ f.(h.v) = −2f.v+ f.(λv) = (λ− 2)f.v.

Next, because V has finite dimension, then there is an eigenspace Vλ that is killed by e. We willcall any v0 ∈ Vλ a maximal vector. Now define vi = 1

i! fi.v0.

We will study the span of vm, . . . , v1, v0, where f.vm = 0 but vm 6= 0.

Example 72. Consider the usual 2-dimensional representation. Then e1 has eigenvalue 1 and e2has eigenvalue −1. In addition, f.e1 = e2 and e.e2 = e1.

For the adjoint representation, f,h, e have weights −2, 0, 1.

We will show that 〈vm, . . . , v0〉 ⊂ V is a subrepresentation:

Proposition 73. The following are true:

1. h.vi = (λ− 2i).vi;

2. f.vi = (i+ 1)vi+1;

3. e.vi = (λ− i+ 1)vi−1.

Proof. We only need to prove the third claim. Here, we note that

e.vi =1i(e.(f.vi−1))

=1i([e, f].vi−1 + f.(e.vi−1))

=1i(h.vi−1 + f.(λ− i+ 2)vi−2)

=1i((λ− 2i+ 2)vi−1 + (i− 1)(λ− i+ 2)vi−1)

= (λ− i+ 1)vi−1.7This was done in Ivan’s homework, and is an exercise in Etingof’s book.

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Corollary 74. If V is irreducible, then Vλ is one-dimensional. In particular, V = 〈vm, . . . , v0〉.

Now all we need to do is classify the highest weights and the dimensions for irreducible represen-tations. Here we observe that

0 = e.(f.vm) = e.(m+ 1)vm+1 = (m+ 1)(λ− (m+ 1) + 1)vm = (m+ 1)(λ−m)vm.

Because m is nonnegative and vm is nonzero, m = λ. Thus we have proved:

Theorem 75. For an irreducible representation V , λ is a nonnegative integer and V has weights λ, λ−2, . . . ,−λ.

Next time we will construct a representation with these weights.

10 Lecture 10 (Oct 3)

10.1 Representation Theory of sl2 Continued Last time, we proved the following result:

Theorem 76. There is at most one irreducible representation of sl2(C) of a given finite dimension. Further-more, its weights are m,m− 2, . . . , 2 −m,−m where m = dimV − 1. In addition, all weight spaces are1-dimensional and there is a unique maximal weight space.

This time, we will show:

Theorem 77. There exists a unique irreducible representation of each positive dimension.

Proof 1. We write down three matrices:

E =

0 n

0 n−1. . . . . .

0 20 1

0

,H =

nn−2

n−4. . .

2−n−n

, F =

01 0

2 0. . . . . .n−1 0

n 0

.

Checking this is left to the reader.8

Proof 2. Let C2 be the standard representation. Then note that (C2)∗ ' C2. Thus we can considerthe infinite-dimensional representation S(V∗) = S(C2).9

Now we can take our representations to be Sk−1(C2). Then we can check that

h.(xiyk−1−i) =∑

xi−1(h.x)yk−1−i +∑k−1−i

xi(h.y)(yk−2−i) = (−k+ 1 + 2i)xiyk−1−i.

Therefore we have the correct weights, so we are done.

Now, Weyl tells us that a general representation of sl2 is a sum of the form

V = ⊕m∈ZaiVi.

To find the multiplicities, just find the largest eigenvalue of h and take away weights of h.8This is taken from Ivan’s homework. Eric normalized the e, so we have a different f.9This was on Ivan’s homework, but for the group SL2.

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10.2 Maximal Tori Consider sln and denote the set of diagonal elements by h. This is an abeliansubalgebra, so we will find an analog for arbitrary semisimple Lie algebras.

Definition 78. A subalgebra of L consisting of semisimple elements is a toral subalgebra..

Theorem 79. Toral subalgebras are abelian.

Proof. Let T be a toral subalgebra and choose x ∈ T . Because x is semisimple, ad x is semisimpleas a map T → T . Thus ad x|T is semisimple.

Choosing an eigenvector y ∈ T , let yi be a basis of eigenvectors for ady. Now we write x =∑ciyi,

we have[y, x] =

∑ci[y,yi] = −ay,

which implies that [x,y] = 0.

We consider the action of H on L. Because adh are commuting semisimple operators, they aresimultaneously diagonalizable.

Definition 80. Define the root space

Lα = x ∈ L | [h, x] = α(h)x for h ∈ H

where α ∈ H∗.

Note that H ⊂ L0. In particular, L0 = CL(H) and

L = CL(H)⊕⊕α6=0

Lα.

Example 81. In sln, [h, eij] = αij(ti − tj)eij. Thus

sln = H⊕⊕i 6=j

Lαij .

We conclude that H is a maximal torus and that H = L0.


Recall our discussion of toral subalgebras from last time.

11.1 Tori in sln

Proposition 82. tr(ad x ady) = 2n tr(xy) for x,y ∈ h.

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Proof. Note ad x has eigenvalues ti − tj and ady has eigenvalues ui − uj. Then

tr(ad x ady) =∑i,j

(ti − tj)(ui − uj)

=∑

tiui +∑

tjuj −∑

tiuj −∑

tjui

= 2n∑

tiui

= 2n tr(xy).

Corollary 83. κ is nondegenerate on H.

Proposition 84. H = CL(H).

Proof. Let x =∑aijeij. Then

[h, x] =∑

aij(ti − tj)eij.

If x /∈ H, then aij 6= 0 for some i 6= j. But then there exiss h ∈ H such that αij(h) = ti − tj.

Theorem 85. In general, H = CL(H) for any maximal torus H in a semisimple Lie algebra L.

11.2 The General Case Take H ⊂ L to be a maximal torus.

Proposition 86. 1. For all α,β ∈ H∗, [Lα,Lβ] ⊂ Lα+β.

2. For x ∈ Lα with α 6= 0, ad x is nilpotent.

3. For α,β ∈ H∗ with α+β 6= 0, Lα ⊥ Lβ.

Proof. 1. [h, [xα, xβ]] = [[h, xα], xβ] + [xα, [h, xβ]] = (α(x) +β(x))[xα, xβ].

2. This follows from finite dimensionality of L.

3. −α(h)κ(x,y) = κ([x,h],y) = κ(x, [h,y]) = β(h)κ(x,y).

Corollary 87. κ is nondegenerate on CL(H).

Proof. If κ(x,y) = 0 for all y ∈ CL(H) = L0, then κ(x, z) = 0 for all z ∈ L. Thus x = 0.

Lemma 88. If x is nilpotent and [x,y] = 0, then κ(x,y) = 0.

Proof. ad x is nilpotent by definition, and because it commutes with ady, ad x ady must also benilpotent.

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Proof of Theorem 85. If x ∈ CL(H), then adX(H) = 0. Therefore ad xs(H) = ad xn(H) = 0, soxs, xn ∈ C.

Now suppose x ∈ CL(H) is semisimple. Then H+ Fx is abelian and is toral because it consists ofcommuting semisimple elements. Thus x ∈ H.

Next we show κ is nondegenerate on H. Suppose κ(h,h ′) = 0 for all h ′ ∈ H. Then choosez ∈ CL(H). Because zn commutes with H, κ(h, zn) = 0 by Lemma 88. Thus h = 0 by Corollary 87.

We now show CL(H) is nilpotent. For z ∈ CL(H), zs ∈ H, so ad zs = 0. On the other hand, ad znis nilpotent. Thus ad z is nilpotent because zs, zn commute. By Engel, CL(H) is nilpotent.

Next we show that H ∩ [CL(H),CL(H)] = 0. Choosing [x,y] ∈ [CL(H),CL(H)] and h ∈ H, wesee κ([x,y],h) = κ(x, [y,h]) = 0, so [CL(H),CL(H)] ⊥ H. Because κ is nondegenerate on H,H∩ [C,C] = 0.

Next note that Z(CL(H))∩ [CL(H),CL(H)] 6= 0. Choose a nonzero z ∈ Z(CL(H))∩ [CL(H),CL(H)].Thus zs, zn ∈ Z(CL(H)). However, [CL(H),CL(H)] is an ideal, so zs ∈ [CL(H),CL(H)]∩H. Thuszs = 0. Therefore κ(z,CL(H)) = 0. Thus z = 0. Therefore CL(H) is abelian.

Finally, we show that H = CL(H). Choose z ∈ CL(H). Then zn ∈ CL(H), so κ(zn,CL(H)) = 0, sozn = 0. Thus z = zs ∈ H.

Corollary 89. κ is nondegenerate on H.

Now we may identify H with H∗ using κ.10 In addition, we may write L = H⊕⊕α6=0 Lα. We

define the set of roots Φ = α ∈ H∗ | Lα 6= 0,α 6= 0..


Last time we proved that κ is nondegenerate on H and that CL(H) = H. Thus we may associatetα ∈ H to every α ∈ H∗.

Proposition 90. 1. Φ spans H∗.

2. If α ∈ H, then −α ∈ Φ.

3. If α ∈ Φ, x ∈ Lα,y ∈ L−α, then [x,y] = κ(x,y)tα.

4. If α ∈ Φ, then [Lα,L−α] is 1-dimensional.

5. α(tα) = κ(tα, tα) 6= 0 for all α ∈ Φ.

6. The span of Lα,L−α, tα is a copy of sl2.

7. hα = 2tακ(tα,tα)

and hα = h−α.

Proof. 1. If not, then thee exists h ∈ H such that α(h) = 0 for all α ∈ Φ. But this impliesh ∈ Z(L), which is a contradiction for nonzero h.

2. Recall that Lα ⊥ Lβ if α+β 6= 0. Thus κ(Lα,L−α) 6= 0, so L−α is nonempty.

10This will help us in our study of root systems.

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3. Just evaluate κ(h, [x,y]). We have

κ(h, [x,y]) = κ([h, x],y)= κ(α(h)x,y)= κ(tα,h)κ(x,y)= κ(κ(x,y)tα,h).

4. Follows from the previous part.

5. The first part follows from the definition. Then if the inner product is zero, [tα, x] = 0 =[y, tα] for all x ∈ Lα,y ∈ L−α. Choosing nonzero x,y such that κ(x,y) 6= 0, rescale soκ(x,y) = 1. Then [x,y] = tα. Then the subalgebra S spanned by x, tα,y is solvable. By Lie’stheorem, [adL S, adL S] consists of nilpotent matrices. Then tα is ad-nilpotent, so it must be0, which is a contradiction.

6. Choose hα a multiple of tα such that [hα, x] = 2x. Now let c be such that hα = ctα andchoose yα such that κ(yα,yα) = c.

7. c2κ(tα, tα) = κ(hα,hα) = cκ(hα, tα) = 2c.

We will call the hα coroots and denote the set of coroots by Φ∨.

Example 91. In sln, α(h) = ti − tj. Then hα = eii − ejj.

Now let Sα be the sl2 associated to α. Then we consider L as an Sα-module. In particular, considerM = H⊕

⊕c Lcα . We can then write M = ker(α)⊕ Sα⊕M ′, where M ′ has all of the odd weights.

Also, this shows that the Lα are one-dimensional.

Now we can also observe that if α ∈ Φ, no other positive multiple of α is in Φ.


Recall our discussion of root systems from last time. We showed that all root spaces are one-dimensional and that if α is a root, no other positive multiple of α is a root. It is easy to showthat:

Proposition 92. 1. [Lα,Lβ] = Lβ+α if α+β is a root.

2. β(hα) ∈ Z and β−β(hα)α ∈ Φ.

3. The Lα generate L.

4. Say β+ qα ∈ Φ but β+ q ′α /∈ Φ for q ′ > q. In addition, suppose β− rα ∈ Φ but β− r ′α /∈ Φfor r ′ > r. Then β+ iα ∈ Φ for all −r 6 i 6 q.

Proof. Consider the space M = ⊕Lβ+iα. Then M is an Sα-module. Then note the eigenvalues ofhα are β(hα) + 2i. Thus β(hα) ∈ Z. In addition, the eigenvalues of hα are distinct and have thesame parity. Thus the last part is true. To complete the second part, note that (β−β(hα)α)(hα) =

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−β(hα). To do the first part, note that ad xα is surjective if and only if β+α is a root (because Mis irreducible). Finally, the third part follows from the fact that the α span H∗.

13.1 An Example: The symplectic Lie algebra Use the standard symplectic form. Then a matrix

X =

(A BC D

)is symplectic if and only if AT = −D,BT = B,CT = C. Thus sp2n has dimension

2n2 +n. We can define H to be the set of diagonal matrices. Then everything simply restricts fromsl2n.

Restricting to sp4, we have four roots: t1 − t2, 2t1, 2t2, t1 + t2. Note that H∗ is an F-vector space.Now consider the Q-vector space EQ spanned by Φ.

Proposition 93. dimQ EQ = dimFH∗.

Proof. Choose an F-basis of H∗ from Φ. Then for β ∈ Φ, we have β =∑ciαi. Then taking the

inner product with β and using that β(hα) ∈ Z, we are done.

Remark 94. The inner product on EQ is positive-definite. Thus we can extend scalars to R to obtaina Euclidean space.


Last time, we defined the real Euclidean space associated to the root system Φ. Now we definethe axioms of a root system.

14.1 Root Systems

Definition 95. A root system is a collection of finitely many nonzero vectors Φ ⊂ E in a Euclideanspace that satisfies:

(R1): Φ spans E;

(R2): If α ∈ Φ and cα ∈ Φ, then c = ±1.

(R3): For α ∈ Φ, σα(Φ) = Φ, where σα is the reflection through α.

(R4): If α,β ∈ Φ, then 〈β,α〉 := 2(β,α)α,α ∈ Z.

Remark 96. If we relax (R2), then our definition is called “reduced.” If we relax (R4), then ourdefinition is called “crystallographic.”

Proposition 97. Φ ⊂ E attached to L is a root system.

Proposition 98. We showed last time that 〈β,α〉 = β(hα) ∈ Z. Also we showed that the root strings areunbroken and that β−β(hα)α is a root. We already knew that the inner product was positive-definite andthat (R1) and (R2) hold.

Example 99. In sln+1, the roots are simply the εi − εj and we use the usual Euclidean structureon Rn induced from gln. Checking the axioms is easy.11

11This was in Ivan’s homework.

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14.2 Classifying Root Systems Note that (R4) limits the possibilities for 〈β,α〉. In addition, wecan analyze the Weyl group. We can also consider 〈β,α〉〈α,β〉 = 4 cos2 θα,β. This constrains theangles between the roots to be π/2,π/6,π/3,π/4, 3π/4, 2π/3, 5π/6, 0,π.

Definition 100. A root system E,Φ is reducible if E = E1 ⊕ E2 and all roots are contained in eitherE1 or E2.

Example 101. If dimE = 1, then we only have one root system, A1. If dimE = 2, then onepossibility is the root system A1 ⊕A1, which is just an orthonormal basis. In addition, we have thecase where α,β have angle 2π/3, which is the A2 root system. If α,β have angle 3π/4, then onehas length

√2 and the other has length 1. This is the root system B2. Finally, if θ = 5π/6, we have

G2.

Definition 102. The Weyl group is the group generated by the σα.

It is easy to see that W ⊂ SΦ. In particular, W is finite.

Example 103. The Weyl group of An is Sn+1.


Recall that the group SLn acts transitively on flags in Cn. Then the stabilizer of any flag is thegroup B of upper triangular matrices. Then the set of all flags can be identified with G/B. Thishas a left action on G/B, so we have

Theorem 104 (Bruhat). The orbits of B on G/B are indexed by the Weyl group.

A standard fact is that the orbit Xw is an affine space of dimension `(w). Here `(w) is the smallestnumber adjacent transpositions needed to write w. In addition, the closure of Xw is no longersmooth, and we can study its singularities, where Kazhdan-Lusztig polynomials appear.12.

15.1 Weyl Groups continued Recall that the Weyl group preserves the inner product. Todaywe will discuss the analogue of the adjacent transpositions that allow us to define the length of anelement in the Weyl group.

Remark 105. Any irreducible root system has a corresponding semisimple Lie algebra.

Definition 106. A set of simple roots ∆ ⊂ Φ is a set of roots such that every root γ ∈ Φ can bewritten γ =

∑∆ cαα, where all cα have the same sign. Also, ∆ must span E.

Example 107. α,α+β form a set of simple roots in A2.

Proposition 108. W acts transitively on the set of bases.

Proposition 109. Every root system has a base.

Proof. Choose γ ∈ E to not be orthogonal to any root. Now define Φ+(γ) to be the set of rootswith positive inner product with γ. Now call β ∈ Φ+ indecomposable if β cannot be written asα1 +α2 for positive roots αi. Now choose ∆(γ) to be the set of indecomposable roots in Φ+(γ).

12The Kazhdan-Lusztig conjectures relate these to representation theory, and were proven by Beilinson and Bernstein

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Now we show that ∆(γ) is a base and every base is of this form. It is easy to see that any positiveroot not in ∆ is decomposable, so then it must be a sum of β1 +β2. Then using the inner productand well-ordering, β must be a nonnegative sum.

Next we show that if α 6= β ∈ ∆(γ), then (α,β 6 0). This is because otherwise, α− β is a root.Now either α−β or β−α is a positive root, which implies that either α or β is decomposible.

Next, we show that ∆(γ) is linearly independent. If not, then we can take the inner product ofsome vector τ with itself to see that τ = 0.

To show that every base is of this form, we can choose γ such that (α,γ) > 0 for all α ∈ ∆.By construction, ∆ ⊂ Φ+(γ). Now note that Φ+ is the set of nonnegative combinations of ∆,so it is contained in Φ+(γ). Also Φ− ⊂ −Φ+(γ). By uniqueness of β =

∑cαα, ∆ consists of

indecomposable elements. Because both ∆,∆(γ) are bases of E, they must be the same.

Now denote α⊥ by Pα. Then the connected components of E \∪Pα are called Weyl chambers.

Theorem 110. Let ∆ be a base of Φ. Then W acts transitively on the set of Weyl chambers and on the setof bases. In addition, each base corresponds to a unique Weyl chamber. In fact, the action of W on the set ofWeyl chambers is faithful.


Let ∆ be a base of Φ. Then define W ′ ⊂W to be the subgroup generated by the simple reflections.Our goal is to show that W ′ =W. Then for any w ∈W ′, we define `(w) to be the smallest t suchthat w = σ1 · · ·σt.

Lemma 111. For α ∈ ∆, σα permutes Φ+ \ α.

Proof. Note that σα only changes the α coefficient of any positive root β. However, this coefficientremains nonnegative.

Lemma 112. Let δ = 12∑α∈Φ+ α. Then 〈δ,α〉 = 1.

Proof. Write δ = 12

(∑β 6=α β+α

), apply σα, and then use the previous lemma.

Lemma 113. Suppose w = σi1 · · ·σit is reduced, then σi1 · · ·σit−1(αt) is a negative root.

Now we define a partial order on Z[Φ] where α 6 β if β− α has positive coefficients in ∆. Wewrite α > 0 if α is a positive root.

Proposition 114. W acts transitively on bases.

Proof. First we show this for W ′. Choose two bases ∆(γ),∆(γ ′). Then note that σ(∆(γ)) = ∆(σ(γ)).Also, σ sends positive roots to positive roots.

Choose σ ∈W ′ such that (σ(γ), δ) is as large as possible. Then for α ∈ ∆, (σασ(γ), δ) 6 (σ(γ), δ).This implies that (σ(γ),α) > 0. Because γ is regular, so is σ(γ). Therefore σ(γ) is in the Weylchamber associated to ∆, as desired.

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Next, any root β is part of some base, so it can be obtained from some simple root by a seriesof simple reglections. Next, we want to show that W ′ = W. This follows from the fact thatgσvg

−1 = σg(v) for any g ∈ O(E).

Finally, we will show that W acts simply transitively on bases. Choosing a reduced expressionfor w, we have w(αit) < 0 if t > 1. However, w(∆) = ∆, so w(α)it > 0. This is a contradiction, sow = 1.

This discussion defines a length function W → Z > 0. Then for w ∈ W, define Φ+(w) = α ∈Φ+ | w(α) < 0. For example, Φ+(1) = ∅ and Φ+(σα) = α.

Proposition 115. There exists a unique w0 ∈W such that Φ+(w0) = Φ+.

Proof. If ∆ is a base, then so is −∆.

Lemma 116. `(w) = #Φ+(w).


The projector is on and the screen is down over the board. I do not know what Eric plans to dowith the computer.

Recall that last time we defined the length function. We need to prove Lemma 116.

Proof of Lemma 116. We use induction. It is true for w = 1 and for w a simple reflection, so thenwe use the fact that the expressions are reduced to write Φ+(w ′) = σit(Φ

+(w) \ αit). Checkingthis simply follows from applying σit to Φ+(w ′).

Lemma 117. Let C be a Weyl chamber for ∆. Then for λ,µ ∈ C, w(λ) = µ only if λ = µ.

Now given ∆, we may associate its Dynkin diagram. To each simple root, we associate a node.There is no connection if two roots are orthogonal, one line if 〈β,α〉, 〈α,β〉 are both −1, two linesif one is −2, and three lines if one is −3. The arrow points to the shorter root.

Example 118. Some Dynkin diagrams:

• An:

• Bn:

• Cn:

• Dn:

• G2:

Now it is easy to see that if Φ is irreducible, then ∆ can not be decomposed in the same way. Thusirreducibility is captured by the Dynkin diagram.

Proposition 119. If Φ is irreducible, there are at most two lengths of roots.

Proposition 120. If Φ is ireducible, there exists a maximal positive root and it is long.

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Next we will classify the irreducible Dynkin diagrams and then show the existence of each possibleLie algebra. For now, we can assume that all roots have the same length (simply laced). We willconsider the affine Dynkin diagram, formed by adjoining −θ to the diagram. Finally, we candecorate the affine diagram by the coefficients of Θ =

∑α∈Φ Kαα. Also, if β1, . . . ,βk are the

nodes connnected to α, we have 2Kα =∑Kβi .

18 Lecture 18 (Nov 5)

Recall the construction of the Dynkin diagrams from last time. Note that Φ is irreducible if andonly if the Dynkin diagram is connected. Our goal is to classify irreducible root systems, which isequivalent to classifying connected Dynkin diagrams. Later, we will see that L is simple if andonly if Φ is irreducible. Today, we will classify simply-laced (ADE) Dynkin diagrams.

Theorem 121. The simply-laced connected Dynkin diagrams are the ADE.

Proof. We will use the highest root θ, which is maximal in the partial order. We know that θ isunique, has positive inner product with all positive roots, and is long. We may adjoin −θ to theDynkin diagram to obtain the affine Dynkin diagram.

By definition, we will attach a coefficient of 1 on −θ. For the extended diagram, we have2ci −

∑i−j cj = 0. We will use this fact to construct the decorated extended diagrams.

Starting from the coefficient of −θ, note that we can either have a degree 2 vertex with label 1,where the other vertices are also 1, or a vertex with 1 attached to a vertex with label 2.

In the first case, it is easy to see that all vertices we attach have label 1, so to have a finite diagram,we must form a cycle. Then we have An.

In the second case, if we branch at the label 2 vertex, it is easy to see that we have Dn. If we donot branch, then we attach a vertex with label 3.

In this case, the next vertex can be either 2 or 4. If we have a 2, then we need a third vertex withlabel 2 and to each vertex of label 2, we attach a 1. This gives us E6. If the fourth vertex is a 4, thenthe next vertex can be 2, 3, 5. In the case of either 2 or 3, we have E7. In the case of a 5, the nextvertex is forced to be 6, which gives us E8.

For non-simply laced root systems, we can show existence by writing down the Dynkin diagrams,writing down a list of vectors in an integral lattice of a given length, and showing that the innerproduct is integral. Then reflecting keeps us in the lattice.

Alternatively, we could use the computer to generate all roots from the Cartan matrix. This givesus a set of roots which is stable under simple reflections. Then this set is also stable under allreflections, and the integrality follows from this.

Proposition 122. If L is simple, thenΦ is irreducible. In ddition, if L is semisimple, then the decompositioninto simples gives an orthogonal decomposition of Φ.

Proof. Suppose Φ = Φ1 ×Φ2 and that α ∈ Φ1,β ∈ Φ2. If α+β is a root, it is in either Φ1 or Φ2.However, α+β has nonzero inner product with both α,β, so α,β commute. Thus the subalgebra

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K generated by Φ1 commutes with β, so K is a proper subset of L. Then K must be an ideal, so Lis not simple.

For the second part, consider the decomposition into irreduciblesLi. Then the maximal torus H inL is simply a direct sum of the Hi.

Corollary 123. Φ is irreducible if and only if L is simple.


We know that each semisimple Lie algebra gives us a Dynkin diagram, but we have not shownwhether each Dynkin diagram gives us a Lie algebra. Also, we need to show that all maximal toriare conjugate and that two Lie algebras are isomorphic iff they have the same root system.

Lemma 124. All maximal Borels of L are conjugate via Aut(L).

We know this to be true for sln by Lie’s theorem.

Lemma 125. In a solvable subalgebra of L, all toral subalgebras are conjugate.

To generate automorphisms of L, we can take x ∈ L nilpotent. Then take exp(ad x), which isan automorphism of L. Then we call the group Int(L) the group generated by exp(ad x) for xnilpotent.

Proposition 126. If σ : L→ L ′ is an isomorphism that takes H to H ′, then it induces an isomorphism ofΦ,Φ ′.

Proof. Let h ∈ H, x ∈ Lα. Then [h, x] = α(h)x = α σ−1(σ(h))σ(x). Therefore σ(x) ∈ Lασ−1 . Thusασ−1 ∈ Φ ′. Then note that σ takes root strings to root strings, so we must have 〈β,α〉 = 〈β ′,α ′〉.Thus σ∗ is an isomorphism of root systems.

Definition 127. H ⊂ L is a Cartan subalgebra if H is nilpotent and NL(H) = H.

Proposition 128. If L is semisimple and H is maximal toral, then H is a Cartan subalgebra.

Proof of this uses the decomposition of L into eigenspaces for H.

Theorem 129. If L is semisimple and H is a Cartan subalgebra, then H is maximal toral.

Define the group E(L) to be the collection of exp(ad x), where ad x is called strongly nilpotent.Here, strongly nilpotent means that x ∈ La(ady) for some y ∈ L and nonzero a.

Theorem 130. If L is solvable, then all Cartans are conjugate under E(L).

Theorem 131. If L is a general Lie algebra, then all Borels are conjugate under E(L).

Corollary 132. If H is maximal toral, then H = CL(s) for some s ∈ L semisimple.

Definition 133. s ∈ L is called regular semisimple if CL(s) has minimal dimension among allsemisimple elements.

Example 134. In sln, the regular semisimple elements are the ones with all distinct eigenvalues.

Corollary 135. All Borels containing Harise from some choice of positive roots Φ+.

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Definition 136. A parabolic subalgebra is a subalgebra containing a Borel.

Theorem 137. All parabolic subalgebra is conjugate under E(L) to a standard parabolic (block upper-triangular).


From a Lie algebra L, we will construct its universal enveloping algebra U(L). This is the left adjointto the Lie functor from associative algebras to Lie algebras.13

Definition 138. A pair (i,U) is a universal enveloping algebra for L if U is an associative algebra overk with a representation i : L→ U satisfying the following condition: for any associative algebra A,if j : L→ A preserves the Lie bracket, then j factors uniquely through i.

Proposition 139. The universal enveloping algebra is unique up to unique isomorphism.

This follows directly from the universal property (equivalently from being the adjoint of the Liefunctor).

Now we will show existence of the universal enveloping algebra. We will construct this by takingthe quotient of the tensor algebra T(V) by the relations [x,y] = x⊗ y− y⊗ x, recalling that thetensor algebra is the left adjoint of the forgetful functor from associative algebras to vector spaces.Note that the universal enveloping algebra is not graded.

Remark 140. Later, we will see that L is isomorphic to its image in U(L) (the unit of the adjunction).

It is easy to see that i([x,y]) = i(x)i(y) − i(y)i(x). This follows directly from the definition of U(L).

Now we will show that (i,U) is universal. Note that the map L→ A factors uniquely through T(L).Then the map from T(L) kills anything of the form x⊗ y− y⊗ x− [x,y], so it factors uniquelythrough U(L).

Now given any left ideal of U(L), we may consider modules U(L)/I. There are also representationsof L. We will use this technique to construct representations of semisimple Lie algebras.

Example 141. Let L = sl2. Then we can construct U(L), which is a filtered algebra.14. The filtrationis given by

Um := π(T0 ⊕ · · · ⊕ Tm).

Then the associated graded algebra is G = U0 ⊕U1/U0 ⊕ · · · with multiplication induced fromU(L). It is easy to see that G is commutative. In addition, it is easy to see that the Casimir elementis xy+ 1

2h2 + yx, which is contained in the center of U(L).

Now we try to construct the two-dimensional representation as a quotient of U(L). Consider theideal I = 〈x,h− 1,y2〉. Then U/I is spanned by 1,y.

13We can just show that the Lie functor preserves limits, but Eric turned down my shitposting request.14The associated graded algebra is a symmetric algebra by the PBW theorem

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Note that by the universal property of the universal enveloping algebra, a representation of L isthe same thing as a representation of U(L). In fact, once we see that L ⊂ U(L), then this gives anequivalence of categories between the two notions.

Theorem 142 (Poincare-Birkhoff-Witt). Choose an ordered basis x1 < x2 < · · · for L. Then the productsxi(1)xi(2) · · · xi(m) where 1 6 i(1) 6 i(2) 6 · · · 6 i(m) form a basis for U(L).

Corollary 143. 1. L ⊂ U(L).

2. If H ⊂ L, then U(H) ⊂ U(L). In addition, U(L) is a free U(H)-module.

Now recall that a semisimple Lie algebra can be decomposed as H⊕N⊕N−, where N is the sumof the weight spaces of the positive roots.

Also recall the symmetric and exterior algebras and their graded pieces. Next, if L is semisimple,∧L splits as an an L-module. We may ask in what degrees the trivial representation appears

because∧m L is an L-module. This allows us to construct a “Frobenius polynomial” by

dimL∑i=1

dim(Hom(k,i∧L))qi.

By a result of Konstant, this polynomial is

dimH∏i=1

(1 + q2mi+1)

where m1 6 · · · 6 mn.

Example 144.∧2 sl2 is the irreducible representation with highest weight 2, so the Frobenius

polynomial is 1 + q3.

Corollary 145. The PBW theorem is equivalent to the following: the associated graded algebra for U(L) isisomorphic to S(L).

After this, we went into a discussion about the infinite-dimensional exercise from the last home-work. Now consider the ideal I ⊂ U(sl2) generated by x,h− λ. Then note U/I is generated by theimage of 1. By construction, x kills 1 and h.1 = λ. Unfortunately, this is infinite-dimensional, so weneed to kill some power of y. By PBW, all powers of y are linearly independent, so U(L)/Iλ ' Z(λ).

Now let L be semisimple with Borel H⊕N. Then N has a basis xα given by the positive rootsΦ+ and H has a basis given by hi for αi ∈ ∆. Then for any λ ∈ H∗, we can define

Iλ = (xα | α ∈ Φ+, hi − λ(hi)).

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Today we will begin the classification finite dimensional irreducible representations of L for Lsemisimple. Let V be a finite dimensional representation of L. Then restricting to H, we obtain that

V =⊕µ∈H∗

Lµ.

We denote the µ-weight space of V by Vµ.

Now recall that the Borel is associated to a set of simple roots. It is easy to see that B is solvable.By Lie’s theorem, there exists v+ ∈ V such that B.v+ ∈ F.v+. In particular, N.v+ = 0 andh.v+ = λ(h).v+ for some λ ∈ H∗.

We will call v+ a maximal vector and λ is called the highest weight. Now we consider what weightsmay occur as highest weights. If β ∈ Φ+, then a choice of nonzero xβ ∈ Lβ we have a uniquehβ ∈ H,yβ∈L−β such that xβ,hβ,yβ is the usual basis of sl2. Now let Si be the sl2 associated tothe simple root αi. Then because v+ is an eigenvector for hi, λ(hi) must be a nonnegative integer.In addition, v+ is maximal for Si.

Definition 146. 1. Any λ ∈ H∗ is called a weight.

2. If 〈λ,αi〉 ∈ Z for all i, then λ is an integral weight.

3. If 〈λ,αi〉 ∈ Z>0, then λ is dominant integral. If 〈λ,αi〉 ∈ Z>0, then λ is strictly dominant.

We will call Λ the set of integral weights and Λ+ the set of dominant integral weights. We focuson the integral weights, which must be rational combinations of the simple roots. We know thatamongst the roots of L there are at most two dominant integral weights: the highest long root Θand the short dominant root Θs.

It is also easy to see that Φ ⊂ Λ, so we have Z[Φ] ⊂ Λ ⊂ E ' Rn. We call Z[Φ] the root latticeand Λ the weight lattice.

Example 147. In A2, the dominant region is 0 < θ < π/3. An example of a dominant weight is13α1 +

23α2.

In addition, the dominant integral root is α1 +α2. Therefore sl3 is a highest weight representationfor Θ = α1 +α2.

The standard representation has maximal vector (1, 0, 0) and highest weight 13 (α2 + 2α1).

This suggests

Proposition 148. If V is a finite dimensional irreducible representation, then λ ∈ Λ+.

First recall that if R is a rint and L is some left ideal, then M ' R/I is a left ideal. In adition, M is acyclic module (generated by one element).

Definition 149. V is a standard cyclic module if V is cyclic and B.v+ = Fv+, where v+ is thehighest weight vector.

Note that any standard cyclic module is of the form U(L)/I, where I contains the span ofxi,hi − λ(hi).

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Theorem 150 (Theorem on standard cyclic modules of highest weight λ). (a) V is spanned by∏β∈Φ+ y

iββ

and V is a direct sum of its weight spaces.

(b) The weights are of the form µ = λ−∑ni=1 Kiαi.

(c) For all µ ∈ H∗, Vµ is finite dimensional and dimVλ = 1.

Proof. Write L = N− ⊕B. Then if u ∈ U(L), by PBW, we have∑I

∏yiIβuI, where β runs over the

positive roots and uI ∈ U(B). Then u.v+ =∑I

∏β y

iIβcI.v

+. Because cI is a constant we have thefirst part of (a).

Next note that v+ ∈ Vλ implies that w has eigenfunctional λ−∑ikβk. This implies (a). For (b),

just write βi as a sum of the αi.

Finally, fix µ ∈ H∗. We need to solve µ = λ−∑ikβk in nonnegative integers. This is the same as

writing λ− µ =∑ikβk. Because the positive roots are nonnegative combinations of simple roots,

there are a finite number of solutions. In addition, there is one solution for λ− µ.

Remark 151. The number of solutions of λ− µ =∑ikβk is called the Kostant partition function.

This is given by ∏β>0

(1 − eβ)−1.

We will prove additional parts of the theorem on standard cyclic modules.

Proposition 152. Each submodule of V is a direct sum of its weight spaces.

Proof. Let W ⊂ V be a submodule. Then w ∈W is given by w = v1 + · · ·+ vn. Then acting by hrepeatedly, we obtain that vn ∈W. Then we use induction.

Proposition 153. V is indecomposable with a unique proper maximal submodule.

Proof. Note that any proper submodule W ⊂ V lies in a sum of weight spaces not containingVλ. Therefore V is indecomposable. In addition, the sum of all proper submodules is the uniquemaximal submodule.

Corollary 154. Standard cyclic modules have a unique irreducible quotient.


Recall that V is standard cyclic if it has a maximal vector which generates V . Now let Φ be anirreducible root system. Then we define the fundamental weights to be the dual basis to 2αi

( αi,αi).We call the final weights λi (or $). Then the λi are a Z-basis of Λ and the Z>0-monoid spannedby the λi is Λ+.

In general, the λi can be written as∑djαj, where 0 < dj ∈ Q. As a consequence, we have

Proposition 155. For λ ∈ Λ+, the number of µ ∈ Λ+ with µ < λ is finite.

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Recall from last time that weight spaces in standard cyclic modules are finite-dimensional andthat V has a maximal proper submodule. Therefore it has a unique irreducible quotient.

It turns out that the standard cyclic module Z(λ) is isomorphic to U(L)⊗U(B)Dλ, where xα.v =0,h.v = λ(h)v. This is called the Verma module of high weight λ. By PBW, we know U(L) is freeover U(B).

It is easy to see that the map from Z(λ) to the Verma module is an isomorphism. Then considerthe unique irreducible quotient V(λ).

Theorem 156. If λ ∈ Λ+, then V(λ) is finite dimensional.

Lemma 157. Any V(λ) has a unique maximal vector.

Proof. If there are two maximal vectors with weights λ,µ, then it is easy to see that λ = µ. We alsoknow that Vλ is one-dimensional.

Lemma 158. 1. [xj,yk+1i ] = 0.

2. [hj,yk+1i ] = −(k+ 1)αi(hj)yk+1

j .

3. [xi,yk+1i ] = −(k+ 1)yki (k− hi).

Proof of Theorem 156. Choose mi = 〈λ,αi〉 ∈ Z>0. First set w = ymi+1i . Then by Lemma 158, we

have xj.w = xi.w = 0, so N.w = 0. Thus w is a maximal vector, but it has weight λ− (mi + 1)αi,so w = 0.

This implies that V(λ) contains a nonzero finite-dimensional Si-module for each i = 1, . . . ,n.Next we observe that V is a sum of finite-dimensional Si-modules. Now let W ⊂ V(λ) bea finite dimensional Si-submodule. Then consider the span of xαW for all α ∈ Φ. This isfinite-dimensional and is an Si-submodule because [xi, xα] is either hi or xα+αi .

Now let S ′ be the sum of finite-dimensional Si-submodules. Then clearly S ′ is L-stable. Thereforebecause V(λ) is irreducible, S ′ = V(λ). Therefore, for each i, V(λ) is a sum of finite-dimensionalSi-submodules. However, each weight space is finite dimensional, so it is contained in some finitesum of finite-dimensional Si-modules. Finaly, the sum⊕

k∈Z

Vµ+kαi

is a finite-dimensional Si-module. Thus we need to show that there are finitely many strings. Bythe standard theory of sl2, we know that the string runs between µ, si(µ), where si is the reflectionacross αi. Also, dimVSi(µ) = dimVµ. Also, there exists w such that w(µ) ∈ Λ+. Therefore allweights in V(λ) are W-conjugate to a dominant weight µ+, so there are only finitely many stringsof the form given above.

24 Lecture 24 (Dec 3)

We begin by finishing the proof of Theorem 156. This is given at the end of the previous lecture.

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Corollary 159. Note that dimV(λ)µ = dimV(λ)w(µ) for w ∈W.

Now it remains to compute the dimension of V(λ)µ and the dimension of V(λ). We have formulasof Weyl, Freudenthal, and Kostant.

Consider G = IntL the group of automorphisms of L generated by eadx for x nilpotent. Thenconstruct the flag variety G/B and the line bundle Lλ = G×B Cλ, where Cλ is given by thecharacter xix−1

j .

Then we have:

Theorem 160 (Borel-Weil-Bott). For any weight λ, we have

H0(G/B,Lλ) =

0 λ /∈ Λ+

V(λ) λ ∈ Λ+.

In addition, if w(λ+ δ) is strictly dominant for some δ,w, then Lλ has a unique nonzero cohomology group

H`(w)(G/B,Lλ) ' V(w(λ+ δ) − δ).

Theorem 161 (Kostant’s Formula). For weights λ,µ, recall that dimZ(λ)µ is the number of ways towrite λ− µ as a nonnegative sum of positive roots. Write p(λ− µ) = dimZ(λ)µ. We have

dimV(λ)µ =∑w∈W

(−1)`(w)p(w(λ+ δ) − (µ+ δ)).


Note that V(λ) is finite dimensional iff λ is dominant integral. Today we will compute dimV(λ)and dimV(λ)µ. First, we will capture the action of H on V . Assume that V is a direct sum ofweight spaces for H (this is always true if V is finite-dimensional). Consider the group ring Z[Λ]where µ ∈ Λ is represented by eµ (so we do not confuse multiplication and addition).

Definition 162. The character of V =⊕µ∈H∗ Vµ is

chV =∑µ∈Λ

(dimVµ)eµ.

Then recall that dimZ(λ)µ = p(λ− µ).

Example 163. The dominant weights (for A2) less than 2α1 + 2α2 are 2α1 +α2,α1 + 2α2,α1 +α2.Then we obtain p(2α1 + 2α2) = 3. More generally, p(aα1 + bα2) = min(a,b) + 1.

Example 164. Note that V(α1 + α2) = L. In the Verma module, we see that α1 + α2,α1,α2,α1 −α2,α2 −α1 have dimension 1, 0,−α1,−α2 have dimension 2, and −α1 −α2 has dimension 3.

Now consider V(2α1 + 2α2). Applying s1, then we obtain −α1 + 2α2, and applying s2, weobtain 2α1 −α2. Then note that 2α1 + 2α2,α1 + 2α2, 2α1 +α2 have dimension 1, and α1 +α2 hasdimension 2, and 0 has dimension 3. Then we put this together to find that dimV(2α1 + 2α2) = 27.

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Recall the notion of a composition series for a finite group. We apply the same notion herefor R-modules over any rings R. In addition, when composition series exist, a version of theJordan-Hölder theorem holds.

Theorem 165. Every Verma module has a composition series. In addition, simple quotients are of the formV(µ), where µ+ δ is w-conjugate to λ+ δ.

Call µ ∼ λ (linked) if there exists w such that w(λ+ δ) = µ+ δ. Then define the set of µ 6 λ thatare linked to λ by Θ(λ). then we have

Corollary 166. chZ(λ) =∑µ∈Θ(λ) cµ ch(V(µ)), where cµ ∈N.

This formula comes from analyzing a short exact sequence. By some Möbius inversion-like process,we obtain ch(V(λ)) =

∑µ∈Θ(λ) d

λµ ch(Z(µ)).


We can define a q-analog of the Konstant partition function:

pq(µ) =∑

(kα∈Sµ)q#nonzero kα.

Then we may deine a q-analog of multiplicity. Then for dominant integral λ,µ, mλµ(q) has nonzerocoefficients and is a Kazhdan-Lusztig polynomial for the affine Weyl group of L.

Sketch of proof of Theorem 165. Consider Z = Z(U(L)). In addition, consider a maximal vector v+

of weight λ. Then h.z.v+ = z.h.v+ = λ(h)z.v+ and xα.z.v+ = z.xα.v+ = 0, so z.v+ is a scalarmultiple of v+. Then this scalar is χλ : Z→ F. In fact, z ∈ Z acts by this scalar on all of Z(λ).

Next choose αi such that 〈λ,αi〉 = m > 0. Then ym+1αi

.v+ is a maximal vector of weightλ− (m+ 1)αi = sαi(λ+ δ) − δ. Then, after taking the quotient, we can obtain that χsiλ = χλ.Thus, χλ is invariant under W.

Now using Harish-Chandra, if Z(λ) is not irreducible, then we have a proper submodule V ⊂ Z(λ).Then choose v a maximal vector of some weight µ. Then we have a morphism Z(µ)→ U(L).v, soχµ = χλ, and thus λ ∼ µ. There are only finitely many µ such that µ ∼ λ, and each Z(λ)µ is finitedimensional, so the process terminates with a composition series.

Theorem 167 (Harish-Chandra). For λ ∈ H∗, if χλ = χµ, then λ ∼ µ.

Remark 168. Z = U(L)G. In addition, by PBW, Z ' S(L)G. Finally, S(L)G ' S(H)W and is apolynomial ring C[f1, . . . , fn]. In addition, |W| =

∏deg fi. We may also define a q-analog∑

w∈Wq`(w) =

∏ qdeg fi − 1q− 1

.

Theorem 169 (Weyl Character Formula). For λ dominant integral, we have

chV(λ) =∑w∈W(−1)`(w)ew(λ+δ)∑w∈W(−1)`(2)ew(δ)

.

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Example 170. For sl2, the Weyl character is

em+1 − e−m−1

e1 − e−1 = em + em−2 + · · ·+ e−m+2 + e−m.

Theorem 171 (Weyl Dimension Formula). Choose an invariant inner product on H∗. Then we have

dimV(λ) =

∏α>0(λ+ δ,α)∏α>0(δ,α)

.

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